A simple pendulum with length 100 cm and bob of mass 250 g is executing S.H.M of amplitude 10 cm. The maximum tension in the string is found to be $$\dfrac{x}{40}$$ N. The value of $$x$$ is ______.

A simple pendulum has length $$L = 100$$ cm $$= 1$$ m, bob mass $$m = 250$$ g $$= 0.25$$ kg, and amplitude $$A = 10$$ cm $$= 0.1$$ m.

The angular frequency is given by the expression $$\omega = \sqrt{\dfrac{g}{L}} = \sqrt{\dfrac{9.8}{1}} = \sqrt{9.8}$$ rad/s.

The corresponding maximum velocity follows from $$v_{\max} = A\omega = 0.1 \times \sqrt{9.8}$$ m/s, which leads to $$v_{\max}^2 = 0.01 \times 9.8 = 0.098$$ m$$^2$$/s$$^2$$.

At the mean position (the lowest point), the tension reaches its maximum value given by $$T_{\max} = mg + \dfrac{mv_{\max}^2}{L}$$; substituting the values yields $$= 0.25 \times 9.8 + \dfrac{0.25 \times 0.098}{1}$$, and hence $$= 2.45 + 0.0245 = 2.4745 \text{ N}$$.

Since the problem states $$T_{\max} = \dfrac{x}{40}$$, one finds $$x = 40 \times T_{\max} = 40 \times 2.4745 = 98.98$$, which upon rounding to the nearest integer gives $$x = 99$$.

Therefore, the value of $$x$$ is $$\boxed{99}$$.